Problem: $h(x) = 6x^{2}-2x$ $g(x) = -x^{2}-3(f(x))$ $f(t) = 7t^{2}-6t+2(h(t))$ $ f(g(0)) = {?} $
Solution: First, let's solve for the value of the inner function, $g(0)$ . Then we'll know what to plug into the outer function. $g(0) = -0^{2}-3(f(0))$ To solve for the value of $g$ , we need to solve for the value of $f(0)$ $f(0) = 7(0^{2})+(-6)(0)+2(h(0))$ To solve for the value of $f$ , we need to solve for the value of $h(0)$ $h(0) = 6(0^{2})+(-2)(0)$ $h(0) = 0$ That means $f(0) = 7(0^{2})+(-6)(0)+(2)(0)$ $f(0) = 0$